The standard deviation of the first n natural | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The standard deviation ($S$.$D$.) is given by the formula: $S.D. = \sqrt{\frac{1}{n} \sum x_i^2 - \left( \frac{1}{n} \sum x_i \right)^2}$.For the

Vedclass · Accepted Answer

$\sqrt{\frac{n^2 - 1}{12}}$

Vedclass · Answer

(D) The standard deviation ($S$.$D$.) is given by the formula: $S.D. = \sqrt{\frac{1}{n} \sum x_i^2 - \left( \frac{1}{n} \sum x_i \right)^2}$.For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$.Substituting these values:$S.D. = \sqrt{\frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} - \left( \frac{1}{n} \cdot \frac{n(n+1)}{2} \right)^2}$$S.D. = \sqrt{\frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}}$$S.D. = \sqrt{\frac{2(n+1)(2n+1) - 3(n+1)^2}{12}}$$S.D. = \sqrt{\frac{(n+1) [2(2n+1) - 3(n+1)]}{12}}$$S.D. = \sqrt{\frac{(n+1) [4n + 2 - 3n - 3]}{12}}$$S.D. = \sqrt{\frac{(n+1)(n-1)}{12}} = \sqrt{\frac{n^2 - 1}{12}}$.

Classes	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Frequency $({f_i})$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

The standard deviation of the first $n$ natural numbers is . . . . . . .

Similar Questions

Calculate the mean,variance,and standard deviation for the following distribution.

Classes $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$

Frequency $({f_i})$ $3$ $7$ $12$ $15$ $8$ $3$ $2$

Let $n$ be an odd natural number such that the variance of $1, 2, 3, 4, \ldots, n$ is $14$. Then $n$ is equal to ..... .

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively,and the mean of the first four observations is $\frac{7}{2}$,then the variance of the first four observations is equal to:

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$. The coefficient of variation is.....$\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module